\newproblem{lay:4_6_29}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.6.29}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Use Exercise 28 to explain why the equation $A\mathbf{x}=\mathbf{b}$ has a solution for all $\mathbf{b}\in\mathbb{R}^m$ if and only if the
	equation $A^T\mathbf{x}=\mathbf{0}$ has only the trivial solution.
}{
  % Solution
	If the equation $A^T\mathbf{x}=\mathbf{0}$ has only the trivial solution, then the dimension of its null space is 0 ($\dim\{\mathrm{Nul}\{A^T\}\}=0$) and 
	by the proposition b of Exercise 4.6.28, $\dim\{\mathrm{Col}\{A\}\}=m$. This means that the columns of $A$ span $\mathbb{R}^m$ and by the Invertible Matrix
	theorem, there is a solution of the equation $A\mathbf{x}=\mathbf{b}$ for all $\mathbf{b}\in\mathbb{R}^m$.
	
	This reasoning could be reversed in all its steps to show that if there is a solution of the equation $A\mathbf{x}=\mathbf{b}$ for all $\mathbf{b}\in\mathbb{R}^m$, then
	the equation $A^T\mathbf{x}=\mathbf{0}$ has only the trivial solution.
}
\useproblem{lay:4_6_29}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
